0523MediumArrayMatrixSimulation

Spinning the Mosaic Layers

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Trains the technique from

LeetCode 1914Cyclically Rotating a Grid

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A tile mosaic is stored as mosaic, a rectangular table of colour codes with an even number of rows and an even number of columns.

The mosaic is made of nested rings. The outermost ring is every cell on the border of the table; the next ring is the border of what is left once that ring is taken away, and so on. Because both side lengths are even, every ring is a closed loop of cells.

Read a ring by starting at its top-left cell and going clockwise: across its top row from left to right, down its right column, back across its bottom row from right to left, then up its left column. One shift moves the code in each position of that reading into the position one place earlier, and the code in the first position moves to the last position.

Every ring is shifted shifts times, all rings at once and each within itself. Return the mosaic afterwards.

Examples

Example 1

Input: mosaic = [[7, 9], [6, 8]], shifts = 1
Output: [[9, 8], [7, 6]]

The whole mosaic is one ring, read clockwise as 7, 9, 8, 6. After one shift each code sits one place earlier in that reading, so the reading becomes 9, 8, 6, 7.

Example 2

Input: mosaic = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]], shifts = 2
Output: [[3, 4, 8, 12], [2, 11, 10, 16], [1, 7, 6, 15], [5, 9, 13, 14]]

The outer ring reads 1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5 and the inner ring reads 6, 7, 11, 10. Each code moves two places earlier in its own reading.

Example 3

Input: mosaic = [[1, 2, 3, 4], [5, 6, 7, 8]], shifts = 7
Output: [[5, 1, 2, 3], [6, 7, 8, 4]]

This mosaic is a single ring, read clockwise as 1, 2, 3, 4, 8, 7, 6, 5. After seven shifts that same reading is 5, 1, 2, 3, 4, 8, 7, 6.

Constraints

rows == mosaic.length
cols == mosaic[i].length
2 <= rows, cols <= 50
rows and cols are both even.
1 <= mosaic[i][j] <= 5000
1 <= shifts <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def spin_mosaic(mosaic: list[list[int]], shifts: int) -> list[list[int]]:

Java

public int[][] spinMosaic(int[][] mosaic, int shifts)

The intended approach

Time O(rows * cols): each cell is read once and written once.Space O(rows * cols) for the returned table, plus O(rows + cols) per ring.

Handle one ring at a time. Walk the ring's cells in the clockwise reading order the statement describes and collect both the coordinates and the codes into two parallel lists, which flattens the ring into a plain sequence. A ring of length L is unchanged by L shifts, so only `shifts % L` of them have any effect; write each code back at the coordinate that many places earlier in the sequence, wrapping round the end. Every cell belongs to exactly one ring, so the whole table is touched a constant number of times.

Pythonentry point: spin_mosaic

def spin_mosaic(mosaic, shifts):
    # O(rows*cols) time: read each layer once into a list, and because a layer of
    # length L returns to itself every L shifts only shifts % L of them matter.
    rows = len(mosaic)
    cols = len(mosaic[0])
    out = [row[:] for row in mosaic]
    for d in range(min(rows, cols) // 2):
        top, left = d, d
        bottom, right = rows - 1 - d, cols - 1 - d
        ring = []
        for c in range(left, right + 1):
            ring.append((top, c))
        for r in range(top + 1, bottom + 1):
            ring.append((r, right))
        for c in range(right - 1, left - 1, -1):
            ring.append((bottom, c))
        for r in range(bottom - 1, top, -1):
            ring.append((r, left))
        vals = [mosaic[r][c] for r, c in ring]
        size = len(vals)
        step = shifts % size
        for i in range(size):
            r, c = ring[i]
            out[r][c] = vals[(i + step) % size]
    return out

The obvious version

Collect the coordinate loop of every ring, then actually carry out the shifts one at a time, on each pass moving every code into the previous position of its loop and wrapping the first code round to the end.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_spin_mosaic(mosaic, shifts):
    rows = len(mosaic)
    cols = len(mosaic[0])
    board = [row[:] for row in mosaic]
    rings = []
    for d in range(min(rows, cols) // 2):
        top, left = d, d
        bottom, right = rows - 1 - d, cols - 1 - d
        ring = []
        for c in range(left, right + 1):
            ring.append((top, c))
        for r in range(top + 1, bottom + 1):
            ring.append((r, right))
        for c in range(right - 1, left - 1, -1):
            ring.append((bottom, c))
        for r in range(bottom - 1, top, -1):
            ring.append((r, left))
        rings.append(ring)
    for _ in range(shifts):
        for ring in rings:
            first = board[ring[0][0]][ring[0][1]]
            for i in range(len(ring) - 1):
                r, c = ring[i]
                nr, nc = ring[i + 1]
                board[r][c] = board[nr][nc]
            r, c = ring[-1]
            board[r][c] = first
    return board

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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